Proof.
Expand $(\omega +\mathrm{dd}^{\mathrm{c}}u)^ n$ by the binomial theorem. The term with $k=n$ integrates to zero, since it is exact:
\[ (\mathrm{dd}^{\mathrm{c}}u)^ n=\mathrm{d}\bigl (\mathrm d^{\mathrm c}u\wedge (\mathrm{dd}^{\mathrm{c}}u)^{n-1}\bigr ). \]
For $1\leq k\leq n-1$, apply lemma 11.1.2 with $f=u$ and $T=(\mathrm{dd}^{\mathrm{c}}u)^{k-1}\wedge \omega ^{n-k}$. Since both $\mathrm{d}$ and $\mathrm d^{\mathrm c}$ annihilate $\mathrm{dd}^{\mathrm{c}}u$,
\[ \mathrm{dd}^{\mathrm{c}}T=(\mathrm{dd}^{\mathrm{c}}u)^{k-1}\wedge \mathrm{dd}^{\mathrm{c}}(\omega ^{n-k}), \]
which gives the $k$th term of (11.1.6). When $n=1$, the sum is empty and the mass is constant.
$\square$
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